Exercise 5.21

Proof. Let $D=21=3\times 7$ ( $D$ is positive, odd and square-free). We first search the $\varphi (D)∕2=6$ integers $a,1\le a\le 21$, such that $(a∕D)=1$.

The first case is equivalent to $a\equiv 1(\mathit{mod}3),a\equiv 1,2,4(mod7)$, that is $a\equiv 1,16,4(\mathit{mod}21)$.

The second case gives $a\equiv -1(\mathit{mod}3),a\equiv -1,-2,-4(mod7)$, that is $a\equiv -1,-16,-4(\mathit{mod}21)$, or equivalently $a\equiv 20,5,17(\mathit{mod}21)$.

So $A=\{1,4,5,16,17,20\}$ is the set of the integers $a$ such that $1\le a\le 21$, $(a∕D)=1$.

As $(21∕3)=(21∕7)=0$, $21$ is not a quadratic residue modulo $3$ or $7$.

$\text{∙}$ $p\equiv 1(\mathit{mod}4)$.

From Ex.5.20, we know that $D=21$ is a quadratic residue modulo an odd prime $p$, $p\ne 3,p\ne 7$, $p\equiv 1(\mathit{mod}4)$, iff $p\equiv a(\mathit{mod}D)$ for some $a\in A$.

$\text{∙}$ $p\equiv -1(\mathit{mod}4)$.

As $D=21\equiv 1(\mathit{mod}4)$, $\left(\frac{D}{p}\right)\left(\frac{p}{D}\right)={(-1)}^{\frac{p-1}{2}\frac{D-1}{2}}=1$, so the same reasoning as in Ex. 5.20 show that $D$ is a quadratic residue modulo $21$ iff $p\equiv a,a\in A$.

Conclusion : $21$ is a quadratic residue for $2$, and for the primes $p$ such that

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